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How to calculate efficiency of Newton's Method

  1. Feb 18, 2015 #1
    Greetings. I was wondering if anyone knew of a way to calculate the efficiency of Newton's Method for a given function:

    I have an equation f(x) and I'm trying to find a value of x = x0 such that f(x0) = 0.

    So I start with a guess x0 and then use that to find a second (usually closer) guess of x1 by using the following:

    x1 = x0 - f(x0)/f'(x0)

    and then a third guess

    x2 = x1 - f(x1)/f'(x1)

    ...and so on until xn = xn-1 +- e, where "e" is some previously defined allowable error.

    In order to find an "efficiency value" of the above method, I'm thinking this has to do with the rate at which f(x) changes versus the rate at which f"(x) changes. So would I take the derivative of f(x) and divide it by the derivative of f'(x)? = f'(x)/f''(x)

    .....or would I take the derivative of the entire construct? = [f(x)/f'(x)]'

    Or are none of these correct?
  2. jcsd
  3. Feb 18, 2015 #2
    Sorry. You can barely see the ' or the '' on those above derivatives, but they are there.
  4. Feb 18, 2015 #3


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    It's not clear what you mean by the 'efficiency' of Newton's method.


    In the article above, it can be shown that under certain conditions, NM has what is called 'quadratic convergence', which means that if NM has settled onto a solution, the error in a subsequent iteration is proportional to the square of the error in the current iteration.
  5. Feb 18, 2015 #4
    That's exactly what I was looking for. Thank you.
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